A rod of length L consists of two halves: one | vedclass

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(D) The rod is divided into two halves,each of length $l = L/2$.Let the axis of rotation pass through the midpoint of the rod.The moment of inertia of

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$\frac{(m_c + m_s)L^2}{12}$

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(D) The rod is divided into two halves,each of length $l = L/2$.Let the axis of rotation pass through the midpoint of the rod.The moment of inertia of the copper half (mass $m_c$,length $L/2$) about the axis passing through its own center of mass is $I_{cm,c} = \frac{1}{12} m_c (L/2)^2 = \frac{m_c L^2}{48}$.The distance of the center of mass of the copper half from the midpoint of the rod is $d = L/4$.Using the parallel axis theorem,the moment of inertia of the copper half about the rod's midpoint is $I_c = I_{cm,c} + m_c d^2 = \frac{m_c L^2}{48} + m_c (L/4)^2 = \frac{m_c L^2}{48} + \frac{m_c L^2}{16} = \frac{m_c L^2 + 3m_c L^2}{48} = \frac{4m_c L^2}{48} = \frac{m_c L^2}{12}$.Similarly,for the silver half (mass $m_s$),the moment of inertia about the rod's midpoint is $I_s = \frac{m_s L^2}{12}$.The total moment of inertia is $I = I_c + I_s = \frac{m_c L^2}{12} + \frac{m_s L^2}{12} = \frac{(m_c + m_s)L^2}{12}$.

$A$ rod of length $L$ consists of two halves: one half is made of copper with mass $m_c$ and the other half is made of silver with mass $m_s$. What is the moment of inertia of the rod about an axis passing through its midpoint and perpendicular to the rod?

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